Question

已知f(n)=n^2(n为正奇数时）f(n)= -n^2(n为正偶数） 若an=f(n)+f(n+1),求Sn

Answer 1

a1=f(1)+f(2)

a2=f(2)+f(3)

a3=f(3)+f(4)

..

an=f(n)+f(n+1)

Sn=f(1)+f(n+1)+2*[(f(n)+f(n-1))+(f(n-2)+f(n-3))+..+(f(3)+f(2))]

n奇数 f(n)+f(n-1)=n^2-(n-1)^2=2n-1

Sn=1-(n+1)^2+2*[(2n-1)+(2n-3)+...+5]=1-(n+1)^2+(2n-1+5)*(2n-1-5)/2=1-(n+1)^2+(2n+4)(n-3)

=1-(n+1)^2+(2n^2-2n-12)

n偶数 f(n)+f(n-1)=-n^2+(n-1)^2=1-2n

Sn=1+(n+1)^2+2*[(1-2n)+(3-2n)+..+5]

=1+(n+1)^2+(6-2n)(-n-2)

=1+(n+1)^2+(2n^2-2n-12)